What are the solutions to the equation 2x^2+x(-x/2)+(-x/2)^2=4 ?

The solutions to the equation 2x^2+x(-x/2)+(-x/2)^2=4 are x= 4/(sqrt(7)),x=-4/(sqrt(7))

Find the zeros of 2x^2+x(-x/2)+(-x/2)^2=4

The zeros of 2x^2+x(-x/2)+(-x/2)^2=4 are x= 4/(sqrt(7)),x=-4/(sqrt(7))

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2 x^{2} + x (- \frac{x}{2}) + (- \frac{x}{2})^{2} = 4

x = \frac{4}{7}, x = - \frac{4}{7}

Decimal Notation

x = 1.51185 \dots, x = - 1.51185 \dots

Solution steps

2 x^{2} + x (- \frac{x}{2}) + (- \frac{x}{2})^{2} = 4

Expand

2 x^{2} + x (- \frac{x}{2}) + (- \frac{x}{2})^{2} : \frac{7 x ^{2}}{4}

\frac{7 x ^{2}}{4} = 4

Move

4

to the left side

\frac{7 x ^{2}}{4} - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot \frac{7}{4} ( - 4 )}{2 \cdot \frac{7}{4}}

0^{2} - 4 \cdot \frac{7}{4} (- 4) = 27

x_{1, 2} = \frac{- 0 \pm 2 7}{2 \cdot \frac{7}{4}}

Separate the solutions

x_{1} = \frac{- 0 + 2 7}{2 \cdot \frac{7}{4}}, x_{2} = \frac{- 0 - 2 7}{2 \cdot \frac{7}{4}}

x = \frac{- 0 + 2 7}{2 \frac{7}{4}} : \frac{4}{7}

x = \frac{- 0 - 2 7}{2 \frac{7}{4}} : - \frac{4}{7}

The solutions to the quadratic equation are:

x = \frac{4}{7}, x = - \frac{4}{7}

3 ln (10) + 2 ln (100)

27 x^{3} - 343 y^{3}

4 y^{2} - 2 y - 12

r^{3} + 8 b^{3}

\frac{x ^{2}}{y ^{2}}

What are the solutions to the equation 2x^2+x(-x/2)+(-x/2)^2=4 ?
The solutions to the equation 2x^2+x(-x/2)+(-x/2)^2=4 are x= 4/(sqrt(7)),x=-4/(sqrt(7))
Find the zeros of 2x^2+x(-x/2)+(-x/2)^2=4
The zeros of 2x^2+x(-x/2)+(-x/2)^2=4 are x= 4/(sqrt(7)),x=-4/(sqrt(7))